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Polymer Solution Thermodynamics: Flory-Huggins Theory of Polymer Solutions

  1. Sep 30, 2014 #1
    1. The problem statement, all variables and given/known data
    (Note: This isn't an assignment problem, more a curiosity about the derivation of an equation - hopefully it is still posted in the right forum..)

    I have working through the derivation for the partial molar Gibbs free energy of mixing from the Flory-Huggins expression for the Gibbs free energy of mixing, however my math skills are (very) rusty - especially when it comes to partial derivatives.

    2. Relevant equations
    Gibbs Free Energy of Mixing (Flory-Huggins Theory):
    ΔGM=R·T[n1·ln(ϕ1)+n2·ln(ϕ2)+n1· 79c40dd2b43f3c03eaf88b5fc4c199b8.png ·ϕ2]

    partial molar Gibbs free energy of dilution:
    Δμ1=R·T[ln(1-ϕ2)+(1-1/r)·ϕ2 + 79c40dd2b43f3c03eaf88b5fc4c199b8.png ·ϕ22]

    where
    KB=Boltzmann Constant,
    ni=number of moles of i,
    ϕi=volume fraction of i,
    79c40dd2b43f3c03eaf88b5fc4c199b8.png =Flory-Huggins Interaction Parameter

    ϕ1=N1/N0=N1/(N1+r·N2)
    ϕ2=(r·N2)/N0=(r·N2)/(N1+r·N2)
    N0=N1+r·N2

    where
    N0 is the number of lattice sites
    N1 is the number of solvent molecules
    N2 is the number of polymer molecules, each occupying "r" lattice sites (or "r" segments)

    and R=KB·NA
    where
    NA= Avogadro's constant

    3. The attempt at a solution
    ΔGM=R·T[n1·ln(ϕ1)+n2·ln(ϕ2)+n1· 79c40dd2b43f3c03eaf88b5fc4c199b8.png ·ϕ2]

    applying R=KB·NA
    ΔGM=KB·T[N1·ln(ϕ1)+N2·ln(ϕ2)+N1· 79c40dd2b43f3c03eaf88b5fc4c199b8.png ·ϕ2]

    expressing ϕ1 and ϕ2 in terms of N1 and N2 gives
    ΔGM=KB·T[N1·ln(N1/(N1+r·N2))+N2·ln((r·N2)/(N1+r·N2))+N1· 79c40dd2b43f3c03eaf88b5fc4c199b8.png ·(r·N2)/(N1+r·N2)]

    Next step would be to take the partial derivative of the equation with respect to N1, I've tried many times but I cannot get anything close to the equation for the partial molar Gibbs free energy of dilution..

    I broke it up, letting
    Ψ1=N1·ln(N1/(N1+r·N2))
    Ψ2=N2·ln((r·N2)/(N1+r·N2))
    Ψ3=N1· 79c40dd2b43f3c03eaf88b5fc4c199b8.png ·(r·N2)/(N1+r·N2)

    taking the partial derivative
    1/dN1=ln(N1/(N1+rN2)+[1-(N1+rN1N2)/(N1+rN2)]
    2/dN1=-N2[(1+rN2)/(N1+rN2)]
    2/dN1= 79c40dd2b43f3c03eaf88b5fc4c199b8.png [(rN2)/(N1+rN2)]+N1 79c40dd2b43f3c03eaf88b5fc4c199b8.png [(-rN2)/(N1+rN2)2

    which when I plug it all back in, gives me a big mess..
     
    Last edited by a moderator: Apr 19, 2017
  2. jcsd
  3. Sep 30, 2014 #2
    Start out with ##\frac{\partial \psi _3}{\partial n_1}##. You need to reduce the rhs to the least common denominator. You almost have it. At least get that term.

    Chet
     
    Last edited by a moderator: May 7, 2017
  4. Sep 30, 2014 #3
    Hey Chet,
    Thanks for the help, I've got the third term now. Am I even close with the other 2 terms? I've been playing around with them a bit more, but I can't quite seem to get it, especially the first term with the ln(1-ϕ2)
     
  5. Oct 1, 2014 #4
    for ##\frac{\partial \psi _2}{\partial N_1}## I'm taking ##\frac{\partial N _2}{\partial N_1}## and ##\frac{\partial ln(rN_2)}{\partial N_1}## as zero based on the assumption that it is similar to taking the derivative of a constant, would this be correct?
     
  6. Oct 1, 2014 #5
    This is correct. But your partial derivatives of psi1 and psi2 with respect to n1 are both incorrect. You need to be more careful. Try again.

    Also note the phi1 = 1 - phi2

    Chet
     
  7. Oct 1, 2014 #6
    Thank you, I've gotten the correct equation now.
     
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